A skier starts from rest at the top of a hill that is inclined at 9.9° with the horizontal. The hillside is 160 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of the snow before coming to rest?
the potential energy the skier had, mg*160Sin9.9 is all converted to heat in friction.
PE= friction on slide+frictiononhorizontal
mg*160sin9.9= mu*mg*Sin9.9*160 + mu*mg*distance.
solve for distance.
To find the distance the skier moves along the horizontal portion of the snow before coming to rest, we need to analyze the forces acting on the skier.
First, let's determine the component of the skier's weight that acts parallel to the slope. This component is given by:
Force_parallel = Weight * sin(θ)
Where:
Weight = mass * acceleration due to gravity
Acceleration due to gravity (g) is approximately 9.8 m/s²
Next, let's determine the frictional force opposing the skier's motion down the slope. The frictional force can be calculated using the equation:
Frictional_force = coefficient of friction * Normal force
Where:
Normal force = Weight * cos(θ)
Now, let's calculate the forces:
1. Weight:
Weight = mass * acceleration due to gravity
2. Normal force:
Normal force = Weight * cos(θ)
3. Force parallel to the slope:
Force_parallel = Weight * sin(θ)
4. Frictional force:
Frictional_force = coefficient of friction * Normal force
Now, we can use these forces to determine the skier's acceleration down the slope using Newton's second law:
Force_parallel - Frictional_force = mass * acceleration
To find the skier's acceleration, we use the equation:
acceleration = (Force_parallel - Frictional_force) / mass
Now that we have the acceleration, we can use the kinematic equation to find the speed of the skier at the bottom of the slope:
v^2 = u^2 + 2as
Where:
v = final speed of the skier (0 m/s since the skier comes to rest)
u = initial speed of the skier (0 m/s since the skier starts from rest)
a = acceleration of the skier (calculated in the previous step)
s = distance traveled along the slope (given as 160 m)
Solving this equation for "a," we have:
0^2 = 0^2 + 2a(160)
Simplifying the equation gives us:
0 = 320a
Therefore, acceleration "a" equals zero, which means the skier does not accelerate or decelerate along the horizontal portion of the snow. As a result, the skier's initial velocity down the slope is the same as the final velocity along the horizontal portion of the snow.
We can now calculate the skier's initial velocity using the equation mentioned above:
v = u + at
Since acceleration is zero, the equation simplifies to:
v = u
This means the skier's initial velocity down the slope is equal to the final velocity along the horizontal portion of the snow.
Therefore, the skier's speed at the bottom of the slope is given by:
v = √(2as)
Solving for "v," we have:
v = √(2 * 9.8 * 160 * sin(9.9°))
Finally, to find the distance the skier moves along the horizontal portion of the snow before coming to rest, we use the equation:
distance = (v^2) / (2 * frictional force)
Substituting the values we have calculated, we can find the answer to the question.